You are studying an enzyme that has a catalytic efficiency of 7x10^7M-1s-1. Knowing that the Km=2.50µM and how much enzyme did you use to obtain a maximum velocity of 2520 µM/min?
To determine the amount of enzyme used to achieve a maximum velocity of 2520 µM/min, we need to calculate the turnover number (kcat) of the enzyme.
The turnover number (kcat) represents the number of substrate molecules converted into product per unit time when the enzyme is saturated with substrate. It is calculated by dividing the catalytic efficiency (kcat/Km) by the substrate concentration.
First, let's convert the given catalytic efficiency from M⁻¹s⁻¹ to s⁻¹:
Catalytic efficiency (kcat/Km) = 7x10^7 M⁻¹s⁻¹
Km = 2.50 µM
To convert µM to M, divide by 10^6:
Km = 2.50 x 10^(-6) M
Now, we can calculate kcat:
kcat = (kcat/Km) * Km
= (7x10^7 M⁻¹s⁻¹) * (2.50 x 10^(-6) M)
= 1.75 s⁻¹
The turnover number (kcat) represents the number of substrate molecules converted per second for each enzyme molecule. In other words, it tells us that each enzyme molecule can convert 1.75 substrate molecules into products every second.
Next, we can calculate the amount of enzyme used to achieve a maximum velocity of 2520 µM/min:
Maximum velocity = kcat * [Enzyme]max
Rearranging the formula:
[Enzyme]max = Maximum velocity / kcat
= 2520 µM/min / 1.75 s⁻¹
To convert µM/min to M/s, divide by 10^6 and 60:
[Enzyme]max = (2520 x 10^(-6) M/min) / (1.75 s⁻¹)
= 1.446 x 10^(-8) M/s
Therefore, you used approximately 1.446 x 10^(-8) M/s of enzyme to obtain a maximum velocity of 2520 µM/min.